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Here is the limit

$$\lim_{x\rightarrow-2} \frac{4x^2 + ax + a + 12}{x^2 + x - 2}.$$

a) Find the constant $a$

b) find the limit

I dont think it is solvable because it didn't tell me as $x \rightarrow -2$, $y \rightarrow $?

user474330
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Secret
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  • $x^2 + x - 2= 0$ for $x =-2$ so the limit does not exist if $(x^2 + ax +a + 12)$ does not equal $0$. So the limit only exists if $4*2^2 +2a + a +12 = 0$ and $(x^2 + x -2)$ is a removable singularity. So sollve for $a$... and find the limit. – fleablood Oct 11 '17 at 00:03
  • This doesn't make sense, as written. Are you trying to determine $$\lim_{x\to-2}\frac{4x^2+ax+a+12}{x^2+x-2},$$ assuming it exists? – Cameron Buie Oct 11 '17 at 00:05

1 Answers1

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I think it's saying that if the limit exists, find $a$ and find the limit.

If $x \to -2$ then the denominator $\to 0$, so in order for the limit to exist, then the numerator must also $\to 0$.

$(4x^2 +ax +a+12) \to (16 -2a +a+12) = (28-a)$ therefore $a = 28$ So that expression becomes:

$$\lim_{x \to -2} \frac{4x^2 + 28x +40}{x^2+x-2} = \lim_{x \to -2} \frac{4(x+2)(x+5)}{(x+2)(x-1)} = -4$$

Hendrata
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